Problem: Divide the following complex numbers. $ \dfrac{-32+8i}{5+3i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${5-3i}$ $ \dfrac{-32+8i}{5+3i} = \dfrac{-32+8i}{5+3i} \cdot \dfrac{{5-3i}}{{5-3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-32+8i) \cdot (5-3i)} {(5+3i) \cdot (5-3i)} = \dfrac{(-32+8i) \cdot (5-3i)} {5^2 - (3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-32+8i) \cdot (5-3i)} {(5)^2 - (3i)^2} = $ $ \dfrac{(-32+8i) \cdot (5-3i)} {25 + 9} = $ $ \dfrac{(-32+8i) \cdot (5-3i)} {34} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-32+8i}) \cdot ({5-3i})} {34} = $ $ \dfrac{{-32} \cdot {5} + {8} \cdot {5 i} + {-32} \cdot {-3 i} + {8} \cdot {-3 i^2}} {34} $ Evaluate each product of two numbers. $ \dfrac{-160 + 40i + 96i - 24 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{-160 + 40i + 96i + 24} {34} = \dfrac{-136 + 136i} {34} = -4+4i $